Predicting the grain size and hardness of AZ91/SiC nanocomposite by artificial neural networks

Abstract

In the present study, SiC nanoparticles were added to as-cast AZ91 magnesium alloy through friction stir processing (FSP) and an AZ91/SiC surface nanocomposite layer was produced. A relation between the FSP parameters and grain size and hardness of nanocomposite using artificial neural network (ANN) was established. Experimental results showed that distribution of nanoparticles in the stirred zone (SZ) was not uniform and SZ was divided into two regions. In the ANN modeling, the inputs included traverse speed, rotational speed, and region types. Outputs were hardness and grain size. The model can be used to predict hardness and grain size as functions of rotational and traverse speeds and region types. To check the adequacy of the ANN model, the linear regression analyses were carried out to compute the correlation coefficients. The calculated results were in good agreement with experimental data. Additionally, a sensitivity analysis was conducted to determine the parametric impact on the model outputs.

Appendix

The computation process is as follows:

• 1. For each hidden neuron, the absolute value of the output layer weight multiply by the absolute value of the hidden layer weight. This process implies for each input. Then, the following tables are obtained:

Weight of various layer
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type	Grain size
1	−55.3178	−95.5384	3.3025	−2.3188
2	−4.1112	4.3033	3.9569	1.3847
3	1.4662	−9.1465	25.9532	1.6541
4	6.6795	25.6189	1.3834	0.17792
5	−113.018	3.3554	5.0194	23.9326
6	−2.4343	6.6717	4.376	1.9267

Weight of various layer
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type	Hardness
1	39.9372	−61.4851	−3.8629	0.092301
2	7.0352	56.0055	15.0031	−0.17464
3	−3.4478	−3.9941	−3.4619	−0.76524
4	0.19705	19.4762	23.4907	−3.3633
5	−1.8155	23.5069	−14.1	−0.99639
6	−7.0957	−0.6522	−4.542	−0.64619

Grain size calculation.
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type	Sum
1	P 11 = 128.2709	P 12 = 221.5344	P 13 = 7.6578	357.4632
2	P 21 = 5.6927	P 22 = 5.9587	P 23 = 5.4791	17.1306
3	P 31 = 2.4251	P 31 = 15.1292	P 33 = 42.9291	60.4836
4	P 41 = 1.1884	P 42 = 4.5581	P 43 = 0.2461	5.9926
5	P 51 = 2704.819	P 52 = 80.3034	P 53 = 120.1273	2,905.25
6	P 61 = 4.6901	P 62 = 12.8543	P 63 = 8.4312	25.97577

Hardness calculation
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type	Sum
1	P 11 = 3.6862	P 12 = 5.6751	P 13 = 0.3565	9.7179
2	P 21 = 1.2286	P 22 = 9.7808	P 23 = 2.6201	13.6295
3	P 31 = 2.6383	P 32 = 3.0564	P 33 = 2.6491	8.3440
4	P 41 = 0.6627	P 42 = 65.5043	P 43 = 79.0062	145.1733
5	P 51 = 1.8089	P 52 = 23.42204	P 53 = 14.0491	39.28009
6	P 61 = 4.58517	P 62 = 0.4214	P 63 = 2.9349	7.9416

2. To obtain Q _ij for each hidden neuron, P _ij was divided into the sum; for all the input variables. For example for 1, ${Q_{{11}}} = {P_{{11}}}/\left( {{P_{{11}}} + {P_{{12}}} + {P_{{13}}}} \right)$.3. For each input neuron, assume S _j to be the sum of Q _ij . For example, ${S_1} = {Q_{{11}}} + {Q_{{21}}} + {Q_{{31}}} + {Q_{{41}}} + {Q_{{51}}} + {Q_{{61}}}$. Then, the following tables are obtained.

Grain size calculation
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type
1	Q 11 = 0.3588	Q 12 = 0.6197	Q 13 = 0.021423
2	Q 21 = 0.33231	Q 22 = 0.347843	Q 23 = 0.319843
3	Q 31 = 0.040097	Q 32 = 0.250137	Q 33 = 0.709765
4	Q 41 = 0.198312	Q 42 = 0.760616	Q 43 = 0.041073
5	Q 51 = 0.931011	Q 52 = 0.02764	Q 53 = 0.041348
6	Q 61 = 0.180559	Q 62 = 0.49486	Q 63 = 0.324581
Sum	S 1 = 2.041131	S 2 = 2.50083	S 3 = 1.458032

Hardness calculation
Hidden layer neuron no.	Traverse speed	Rotational speed	Region type
1	Q 11 = 0.379324	Q 12 = 0.583986	Q 13 = 0.03669
2	Q 21 = 0.090144	Q 22 = 0.717616	Q 23 = 0.192239
3	Q 31 = 0.316202	Q 32 = 0.366303	Q 33 = 0.317495
4	Q 41 = 0.004565	Q 42 = 0.451214	Q 43 = 0.54422
5	Q 51 = 0.046052	Q 52 = 0.596283	Q 53 = 0.357665
6	Q 61 = 0.57736	Q 62 = 0.053068	Q 63 = 0.369572
Sum	S 1 = 1.413648	S 2 = 2.768471	S 3 = 1.817881

4. By dividing each S _j into the sum, the relative importance of each input parameter can be calculated. For example, for the first input parameter (traverse speed), the relative importance is equal to $\left( {{S_1} \times 100} \right)/\left( {{S_1} + {S_2} + {S_3}} \right)$.

Relative importance of input parameters on grain size.
Relative importance (%)
Traverse speed	Rotational speed	Region type
34.01885	41.68061	24.30054
Relative importance of input parameters on hardness.
Relative importance (%)
Travers speed	Rotational speed	Region type
23.5608	46.14119	30.29802